SGU438 the glorious Karlutka river =) (maximum flow)

The topic probably said that there is m person to cross a wide W river, the longest long jump distance is D, the river has n garbage heap, each rubbish heap has coordinates and the same time can accommodate the number of people, ask everyone to jump at least several times to jump to the opposite shore.

It's another question. The maximum flow based on time split.

Two minutes to build the capacity of the network to determine: According to the time of each garbage heap demolition, and then split into two points in the middle of the capacity for the same time can accommodate the number of sides, all T-moment points to all the t+1 moment can reach the point of the edge.

In addition, it can be known that if you can reach the other side of the worst case of total need to jump n+m. Because the worst case of each garbage heap can only accommodate one person, and must jump all the rubbish heap to reach the other side, then the first person needs to jump m+1 times, the back n-1 people follow the front that person jump, each jump once there is a new person to reach the other side, but also need n-1 times, a total of n+m times.

1#include <cstdio>2#include <cstring>3#include <queue>4#include <algorithm>5 using namespacestd;6 #defineINF (1&LT;&LT;30)7 #defineMAXN 111118 #defineMAXM 6666669 Ten structedge{ One     intV,cap,flow,next; A }EDGE[MAXM]; - intVs,vt,ne,nv; - intHEAD[MAXN]; the  - voidAddedge (intUintVintcap) { -Edge[ne].v=v; Edge[ne].cap=cap; edge[ne].flow=0; -Edge[ne].next=head[u]; head[u]=ne++; +Edge[ne].v=u; edge[ne].cap=0; edge[ne].flow=0; -EDGE[NE].NEXT=HEAD[V]; head[v]=ne++; + } A  at intLEVEL[MAXN]; - intGAP[MAXN]; - voidBFs () { -memset (level,-1,sizeof(level)); -Memset (Gap,0,sizeof(GAP)); -level[vt]=0; ingap[level[vt]]++; -queue<int>que; to Que.push (VT); +      while(!Que.empty ()) { -         intu=Que.front (); Que.pop (); the          for(intI=head[u]; i!=-1; I=Edge[i].next) { *             intv=edge[i].v; $             if(level[v]!=-1)Continue;Panax Notoginsenglevel[v]=level[u]+1; -gap[level[v]]++; the Que.push (v); +         } A     } the } +  - intPRE[MAXN]; $ intCUR[MAXN]; $ intIsap () { - BFS (); -memset (pre,-1,sizeof(pre)); thememcpy (Cur,head,sizeof(head)); -     intu=pre[vs]=vs,flow=0, aug=INF;Wuyigap[0]=NV; the      while(level[vs]<NV) { -         BOOLflag=false; Wu          for(int&i=cur[u]; i!=-1; I=Edge[i].next) { -             intv=edge[i].v; About             if(Edge[i].cap!=edge[i].flow && level[u]==level[v]+1){ $flag=true; -pre[v]=u; -u=v; -                 //aug= (Aug==-1?edge[i].cap:min (Aug,edge[i].cap)); AAug=min (aug,edge[i].cap-edge[i].flow); +                 if(v==VT) { theflow+=; -                      for(u=pre[v]; V!=vs; v=u,u=Pre[u]) { $edge[cur[u]].flow+=; theedge[cur[u]^1].flow-=; the                     } the                     //Aug=-1; theaug=INF; -                 } in                  Break; the             } the         } About         if(flag)Continue; the         intMinlevel=NV; the          for(intI=head[u]; i!=-1; I=Edge[i].next) { the             intv=edge[i].v; +             if(Edge[i].cap!=edge[i].flow && level[v]<minlevel) { -Minlevel=Level[v]; thecur[u]=i;Bayi             } the         } the         if(--gap[level[u]]==0) Break; -level[u]=minlevel+1; -gap[level[u]]++; theu=Pre[u]; the     } the     returnflow; the } - intx[ -],y[ -],c[ -]; the intn,m,d,w; the BOOLisOKintTime ) { thevs=n*time*2+1; vt=vs+1; nv=vt+1; Ne=0;94memset (head,-1,sizeof(head)); theAddedge (vs,n*time*2, m); the      for(intI=0; i<n; ++i) { the         if(y[i]<=d) {98              for(intt=0; t<time; ++t) Addedge (n*time*2, i*time+t,m); About         } -         if(w-y[i]<=d) {101              for(intt=0; t<time; ++t) Addedge (i*time+t+n*time,vt,m);102         }103     }104      for(intI=0; i<n; ++i) { the          for(intt=0; t<time; ++t) Addedge (i*time+t,i*time+t+n*time,c[i]);106          for(intt=0; t<time-1; ++t) {107Addedge (i*time+t+n*time,i*time+t+1, m);108              for(intj=0; j<n; ++j) {109                 if(I==j | | y[i]>y[j])Continue; the                 if((X[i]-x[j]) * (X[i]-x[j]) + (Y[i]-y[j]) * (Y[i]-y[j]) <=d*d) Addedge (i*time+t+n*time,j*time+t+1, m);111             } the         }113     } the     returnISAP () = =m; the } the intMain () {117scanf"%d%d%d%d",&n,&m,&d,&W);118      for(intI=0; i<n; ++i) {119scanf"%d%d%d", x+i,y+i,c+i); -     }121     if(d>=W) {122Putchar ('1');123         return 0;124     } the     intL=1, r=n+m+1;126      while(l<R) {127         intMid=l+r>>1; -         if(isOK (mid)) r=mid;129         ElseL=mid+1; the     }131     if(l==n+m+1) puts ("Impossible"); the     Elseprintf"%d", L +1);133     return 0;134}

SGU438 the glorious Karlutka river =) (maximum flow)